A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.

Question

A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at an angle of ? = 39? with the boom, is attached at a distance of 2.38 m from the hinge at the wall. The weight of the sign is 103 N.

What is the tension in the cable and what are the horizontal and vertical forces Fx and Fy exerted on the boom by the hinge? Assume that the +x direction is to the right and the +y direction is up.

T=_____N

Fx=____N

Fy=____N

A sign is supported by a uniform horizontal boom of length 3.00 m and weight 80.0 N. A cable, inclined at an angle of ? = 39? with the boom, is attached at a distance of 2.38 m from the hinge at the wall. The weight of the sign is 103 N. What is the tension in the cable and what are the horizontal and vertical forces Fx and Fy exerted on the boom by the hinge? Assume that the +x direction is to the right and the +y direction is up. T=_____N Fx=____N Fy=____N I tried different formulas, but i keep getting a very wide range of answers. thank you

Explanation / Answer

Equate moment about hine to aero Tsin39*2.38-80*1.5-103*3=0 T=286.423 N Balancing X-direction forces Fx=TcosA =286.423cos39=222.593 N to right Balancing forces in y-direction 286.423sin39-80-103=Fy Fy=-2.748 N Fy=2.748 N downwards

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